" c) "x^(2)+3x+2" by "x+1

If g(x) = x^2 +x -2 and 1/2 (gof) (x) = 2x^2 - 5x +2 , then f(x) is equal to a) 2x -3 b) 2x+3 c) 2x^2 +3x +1 d) 2x^2 -3x -1

The equations ax^(2) + bx+ c =0, x^(3) -2x^(2) +2x -1 =0 have tow roots common, then find the value of a+b.

The anti derivative of sqrt(x)+1/sqrt(x) is a) 1/3 x^(1/3)+2 x^(1/2)+C b) 2/3 x^(2/3)+1/2 x^2+C c) 2/3 x^(3/2)+2 x^(1/2)+C d) 3/2 x^(3/2)+1/2 x^(1/2)+C

int (2x ^ (3) + 3x ^ (2) + 4x + 5) / (2x + 1) dx equls to: (A) (x ^ (3)) / (2) + (x ^ (2)) / (2) +3 (x) / (2) + (7) / (4) ln (2x + 1) + C (B) 5 (x ^ (3)) / (2) + 6 (x) / (2) + (7) / (4) ln (2x + 1) + C (C) 3 (x ^ (3)) / (2) + 3 (x ^ (2)) / (2) + (7 ) / (4) ln (2x + 1) + C (D) (x ^ (2)) / (2) + 3 (x) / (2) + (7) / (4) ln (2x + 1) + C

int(x+1)/(x^(1//2)) dx= a) -x^(3//2)+x^(1//2)+c b) x^(1//2 c) 2/3x^(3//2)+2x^(1//2)+c d) x^(3//2)+x^(1//2)+c

(3x-2)/((x +1) ^(2) (x +2))=(A)/(x+1) + (B)/((x +1)^(2))+ (C )/(x +3), then A + B+C =

If (3x+2)/((x+1)(2x^(2)+3))=A/(x+1)+(Bx+C)/(2x^(2)+3), " then "A+C-B=

int frac {x+1}{x+2} dx =............ A) x+2 - log(x+2) + c B) sqrt (x^2 + 3x +2) -(1/2) log x + (3/2) + sqrt (x^2 + 3x +2) + c C) sqrt (x^2 + 3x +2) +(1/2) log x + (3/2) + sqrt (x^2 + 3x +2) + c D) sqrt (x^2 + 3x +2) -(1/2) log x + (3/2) - sqrt (x^2 + 3x +2) + c